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EDITED FOR (hopeful) CLARITY:

For a simple Lie algebra $\mathfrak{g}$ (over $\mathbb{C}$) and its adjoint group G, the $G$-invariant polynomials on $\mathfrak{g}$ are linear combinations of products of traces, i.e. linear combinations of products of functions of the form $$P(X) = tr_V(\pi(X)^n)$$ for representations $(V,\pi)$, by the Chevalley Restriction theorem.

Is this true for tensors as well? In the following sense: We define functions $$T_V(X_1,\ldots,X_n) = tr_V(\pi(X_1)\cdots\pi(X_n))$$ for representations $(V,\pi)$. For an integer $n$, a permutation $\sigma \in S_n$, a sequence of integers $1 = k_0 < k_1 < \ldots k_m = n$, and a sequence of representations $(V_i,\pi_i)$, we define a tensor monomial as $$\prod_{i=1}^m T_{V_i}(X_{\sigma(k_{i-1}+1)},X_{\sigma(k_{i-1}+2)},\ldots,X_{\sigma(k_i}))$$ The purpose of the $\sigma$ is to distribute the variables across the various traces, so that both $$T(X_1,X_2)T(X_3,X_4,X_5)$$ and $$T(X_2,X_4)T(X_3,X_1,X_5)$$ are tensor monomials in this sense.
Question: Is it true that all of the $G$-invariant elements of $\bigotimes^* \mathfrak{g}^*$ are linear combinations of tensor monomials?

Note: I am only interested in tensor invariants of the adjoint representation; the tensor invariants of the minimal representations have been worked out to various extents, but those are not the ones I'm directly interested in.

I have managed to prove this for the classical Lie algebras, and indeed for $A_r, B_r$ and $C_r$ we can take the minimal representation for $(V,\sigma)$. For $D_r$ we need to do something for the Pfaffian, but once we have that we have everything. All of this follows from the invariant theory of these Lie algebras, which is fairly straightforward.

Unfortunately, the invariant theory of the exceptional Lie algebras is somewhat more complicated, and so I haven't been able to show this statement for them. I think it's true, and I have some little bits of computational evidence that we only have to use the minimal representations for these cases as well, but no proof. Is there any literature on this?

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  • $\begingroup$ Perhaps I am not understanding something. The invariant tensors $T$ that you define by the above displayed formula must satisfy $$T(X_1,X_2,\ldots,X_n)=T(X_2,X_3,\ldots,X_n,X_1)$$ for any $X_1,\ldots,X_n$, but the nonzero exterior forms of degree $n=2m$ obviously don't satisfy this. Thus, for example, $A_2$ has an invariant alternating form of degree $8$, so it can't be made by the above formula for any $(V,\pi)$. Perhaps you should explain more fully what you mean by 'along with rearrangement of the tensor indices'. $\endgroup$ Commented Jun 29, 2014 at 6:22
  • $\begingroup$ I meant to include products of such traces as well. So we include things such as $T(X_1,X_2,X_3)T(X_4,X_5,X_6,X_7,X_8)-T(X_2,X_3,X_4)T(X_5,X_6,X_7,X_8,X_1)+....$ which can be made to alternate. By rearrangement of indices, I tend to think of this in terms of Einstein notation, so I'm thinking $T_{a_1,a_2,...}$ and I want to be able to say that both $T_{a_1,a_2}T_{b_1,b_2,b_3}$ and $T_{a_1,b_1}T_{a_2,b_2,b_3}$ are tensor products of the same traces, differing by how the indices are assigned. $\endgroup$ Commented Jun 30, 2014 at 1:29

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